ChR ISM
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ChR ISM


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( )2
2
2
1 cv
mcmcK
\u2212
=+ 
 
Solving for ( )21 cv\u2212 yields: 
 ( )
K
mc
K
mc
cv 2
2
2
1
1
+
=\u2212 (2) 
 
Evaluate 
K
mc2 to obtain: 
 
( )( )
11048.5
GeV 1035.3
MeV/amu 932amu 197
3
4
2
<<×=
×=
\u2212
K
mc
 
 
Because mc2 << K, equation (2) can 
be approximated by: 
 
( )
K
mccv
2
21 \u2248\u2212 (3) 
Solve for v/c to obtain: 
 
22
1 \u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212=
K
mc
c
v 
 
Chapter R 
 
 
1086 
Expanding the radical binomially 
yields: 
 
sorder termhigher 
2
111
2222 2
1
+
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212=\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212
K
mc
K
mc
 
 
Because mc2 << K: 
 
2222
2
111
2
1
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212\u2248\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212
K
mc
K
mc 
and 
\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212=
22
2
11
K
mccv 
 
Substituting for v in equation (1) 
yields: 
c
K
mc
K
mcccv
22
22
2
1
2
11\u394
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b=
\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212\u2212=
 
 
Substitute numerical values and evaluate \u394v: 
 
( ) km/s .504m/s 10998.2
GeV 1035.3
MeV/amu 932amu 197
2
1
\u394 8
2
4 =×\u239f\u23a0
\u239e\u239c\u239d
\u239b
×
×=v 
 
(b) Because the nuclei are, to three 
significant figures, traveling at the 
speed of light: 
s 334.0
m/s102.998
m 100\u394
\u394 8 \u3bc=×=\u2248 c
xt 
 
 
56 \u2022\u2022\u2022 Consider the flight of a beam of neutrons produced in a nuclear 
reactor. These neutrons have kinetic energies of up to 1.00 MeV. The rest energy 
of a neutron is 939 MeV. (a) What is the speed of 1.00 MeV neutrons? Express 
your answer in terms of v/c. (b) If the average lifetime of such a neutron is 15.0 
min, what is the maximum length of a beam of such neutrons (in a vacuum, in 
the absence of any interactions between the neutrons and other material)? 
Estimate this maximum range by calculating the length corresponding to five 
mean lifetimes. After five mean lifetimes only e\u22125 or 0.007 (0.7%) of the 
neutrons remain. (c) Compare this to the range of so-called \u2033thermally 
moderated\u2033 neutrons, whose kinetic energies are around 0.025 eV. Express your 
answer as a percentage. That is, what percent of the 1.00 MeV-neutron range is 
the thermally moderated-neutron range? (Note that our assumption of a vacuum 
continues, however in reality neutrons of this energy interact readily with matter, 
such as air or water, and \u2033real\u2033 ranges are much shorter.) 
 
Special Relativity 
 
 
1087
Picture the Problem We can apply ( )222 1 cvmcmcK \u2212=+ , the expression 
for the total relativistic energy of a particle, to find the speed of 1.00 MeV 
neutrons. We can use the equation relating distance, rate, and time to estimate 
their maximum beam length. Finally, we can use the non-relativistic expression 
for their kinetic energy to find the range of \u2033thermally moderated\u2033 neutrons. 
 
(a) The total relativistic energy of a 
neutron is given by Equation R-15: 
 
( )2
2
2
1 cv
mcmcK
\u2212
=+ 
 
 
Solving for v/c yields: 
 
2
21
11
\u239f\u239f
\u239f\u239f
\u23a0
\u239e
\u239c\u239c
\u239c\u239c
\u239d
\u239b
+
\u2212=
mc
Kc
v 
 
Substitute numerical values and 
evaluate v/c: 
%61.4
MeV 939
MeV 00.11
11
2
=
\u239f\u239f
\u239f\u239f
\u23a0
\u239e
\u239c\u239c
\u239c\u239c
\u239d
\u239b
+
\u2212=
c
v 
 
(b) Express the maximum expected 
beam length of the neutrons as a 
function of the number n of 
lifetimes: 
 
tc
c
vnx \u394\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b= (1) 
Evaluate this expression for n = 5: 
 
( )( )( ) m 106.22
min
s 60min 0.15m/s 102.9980461.05\u394 108 ×=\u239f\u23a0
\u239e\u239c\u239d
\u239b ××=x 
 
(c) Express the ratio r of the range of 
\u2033thermally moderated\u2033 neutrons to 
the range of 1.00 MeV neutrons: 
 
MeV 1.00
moderated thermally 
\u394
\u394
x
x
r = (2) 
Because the energy of the \u2033thermally 
moderated\u2033 neutrons is 0.025 eV, we 
can use the non-relativistic equation 
for their kinetic energy: 
 
2
2
2
12
2
1 c
c
vmmvK \u239f\u23a0
\u239e\u239c\u239d
\u239b== \u21d2 22mc
K
c
v = 
Substitute for v/c in equation (1) to 
obtain: tcmc
Knx \u3942\u394 2moderated thermally = 
Chapter R 
 
 
1088 
Substitute numerical values and evaluate moderated thermally \u394x : 
 
( ) ( )( ) m 1084.9min 0.15m/s 102.998
MeV 939
eV 025.025\u394 68moderated thermally ×=×=x 
 
Substitute numerical values in 
equation (2) and evaluate r: %016.0m1022.6
m 1084.9
10
6
=×
×=r 
 
57 \u2022\u2022\u2022 You and Ernie are trying to fit a 15-ft-long ladder into a 10-ft-long 
shed with doors at each end. You suggest to Ernie that you open the front door to 
the shed and have him run toward it with the ladder at a speed such that the 
length contraction of the ladder shortens it enough so that it fits in the shed. As 
soon as the back end of the ladder passes through the door, you will slam it shut. 
(a) What is the minimum speed at which Ernie must run to fit the ladder into the 
shed? Express it as a fraction of the speed of light. (b) As Ernie runs toward the 
shed at a speed of 0.866c, he realizes that in the reference frame of the ladder, it 
is the shed which is shorter, not the ladder. How long is the shed in the rest frame 
of the ladder? (c) In the reference frame of the ladder is there any instant that 
both ends of the ladder are simultaneously inside the shed? Examine this from 
the point of view of relativistic simultaneity. 
 
Picture the Problem Let the letter \u2033L\u2033 denote the ladder and the letter \u2033S\u2033 the 
shed. We can apply the length contraction equation to the determination of the 
minimum speed at which Ernie must run to fit the ladder into the shed as well as 
the length of the shed in rest frame of the ladder. 
 
(a) Express the length LL of the shed 
in Ernie\u2019s frame of reference in 
terms of its proper length LL,0: 
 
2
0L,L 1 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vLL \u21d2 2
L,0
2
L1
L
Lcv \u2212= 
 
Substitute numerical values and 
evaluate v: 
 
ccv 75.0
ft15
ft101
2
=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212= 
 
(b) Express the length LS of the shed 
in the rest frame of the ladder in 
terms of its proper length LL,0: 
 
2
2
0S,S 1 c
vLL \u2212=
 
 
Substitute numerical values and 
evaluate LS: 
 
( ) ft0.5866.01ft10 2S =\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
cL 
 
(c) No. In your rest frame, the back end of the ladder will clear the door before the 
front end hits the wall of the shed, while in Ernie\u2019s rest frame, the front end will 
Special Relativity 
 
 
1089
hit the wall of the shed while the back end has yet to clear the door. 
 
Let the ladder be traveling from left to right. To \u2033explain\u2033 the simultaneity issue 
we first describe the situation in the reference frame of the shed. In this frame the 
ladder has length LL = 7.5 m and the shed has length LS,0 = 10 m. We (mentally) 
put a clock at each end of the shed. Let both clocks read zero at the instant the 
left end of the ladder enters the shed. At this instant the right end of the ladder is 
a distance LS, 0 \u2212 LL = 2.5 m from the right end of the shed. At the instant the 
right end of the ladder exits the shed both clocks read \u394t, where ( ) ns63.9\u394 LS,0 =\u2212= vLLt . There are two space-time coincidences to consider: 
the left end of the ladder enters the shed and the clock at the left end of the shed 
reads zero, and the right end of the ladder exits the shed and the clock on the right 
end of the shed reads ( ) ns63.9LS,0 =\u2212 vLL . In the reference frame of the ladder 
the two clocks are moving to the left at speed v = 0.866c. In this frame the clock 
on the right (the trailing clock) is ahead of the clock on the left by 
ns, 9.282S,0 =cvL so when the clock on the right reads 9.63 ns the one on the left 
reads \u201319.3 ns. This means the left end of the ladder is yet to enter the shed when 
the right end of the ladder is exiting the shed. This is consistent with the assertion 
that in the rest frame of the ladder, the ladder is longer than the shed, so the entire 
ladder is never entirely inside the shed. 
 
 
 
 
 
 
 
 
 
Chapter R 
 
 
1090